Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 51526 by ajfour last updated on 27/Dec/18

Commented by ajfour last updated on 27/Dec/18

$$Fortheareainblackandyellow$$$$tobethesame,findrintermsofR.$$$$$$

Answered by mr W last updated on 28/Dec/18

$$\cos \theta =\frac{R-r}{R+r}=\frac{1-\frac{r}{R}}{1+\frac{r}{R}}=\frac{1-\lambda }{1+\lambda }$$$$\sin \theta =\frac{\sqrt{{(1+\lambda )}^2-{(1-\lambda )}^2}}{1+\lambda }=\frac{2\sqrt{\lambda }}{1+\lambda }$$$$$$$$A_{black}=\frac{(R+r)(R+r)\sin \theta }{2}-\frac{R^2\theta }{2}-\frac{r^2}{2}(\pi -\theta )$$$$A_{black}=\frac{{(R+r)}^2\sin \theta }{2}-\frac{(R^2-r^2)\theta }{2}-\frac{\pi r^2}{2}=\pi r^2$$$${(R+r)}^2\sin \theta -(R^2-r^2)\theta =3\pi r^2$$$$\Rightarrow 2(1+\lambda )\sqrt{\lambda }-(1-{\lambda }^2){\cos }^{-1}\frac{1-\lambda }{1+\lambda }=3\pi {\lambda }^2$$$$\Rightarrow \lambda =0.0886$$$$\Rightarrow r=0.0886R$$

Commented by mr W last updated on 28/Dec/18

Commented by ajfour last updated on 28/Dec/18

$$yesSir,ihadnotcaculatedbut$$$$hadarrivedatthesameequation.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com